How do I integrate 1/(1+e^x)?

dextercioby · Sep 9, 2007

Are you differentiating [itex]\ln(u^2 -u)[/itex] wrt "x" or to "u" ?

PowerIso · Sep 9, 2007

opps I meant to type d/du

Hurkyl · Sep 9, 2007

camilus said:

[tex]\int{1 \over u^2-u}du = ln(u^2-u) + C[/tex]

Why would you think that?

p.s. have you looked at your private messages?

camilus · Sep 9, 2007

http://i159.photobucket.com/albums/t121/camilus23/etothexproblem.jpg

Summary:

[tex]\int {1 \over e^x +1}dx = ln \left|{e^x \over e^x+1}\right| + C[/tex]

asd1249jf · Sep 10, 2007

Lol jesus how many mathematicians does it take to solve an exponential integral problem?

I'm not calling anyone stupid, but just found it funny ya'll are being cooperative to solve this

Hurkyl · Sep 10, 2007

Solve a man's integral problem, and he gets one answer. Teach a man how to solve integral problems, and he gets all of his homework done.

(and without cheating)

asd1249jf · Sep 10, 2007

Hurkyl said:

Solve a man's integral problem, and he gets one answer. Teach a man how to solve integral problems, and he gets all of his homework done. (and without cheating)

(APPLAUSE) Well Said!

How do I integrate 1/(1+e^x)?

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